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100027 求三角形各边各角度已知两边和第三边所夹两角

时间:2022-12-04 18:00:46浏览次数:39  
标签:cos 30 40 arccos 已知 各边 100027 三角形 两角

<?php
header('Content-Type: text/html; charset=utf-8');
define ('ROOT', $_SERVER['DOCUMENT_ROOT']);
include ROOT.'/assets/php/head.php';

$tit= '求三角形各边各角度已知两边和第三边所夹两角';
/**
 * a为30,b为40,ao为42,bo为63
 * a为30,b为40,ao为37,bo为53
 * a为30,b为40,ao为25,bo为35
 */
$val='30 40 25 35';
$img='';

//调用方法
mill($val,$img,$tit);

/**
 * mill 是磨粉机的方法
 * $val 传值过来计算,以空格分割成数值
 * $img 自定义图片名,默认以文件名为图片名
 * $tit 标题名
 * */
function mill($val,$img,$tit){
    //初始化
    include ROOT.'/assets/php/init.php';
    imgt($img,$tit);

    //设置小数点保留位数 
    bcscale (2);
    //$pi = round(pi(),2);

     //以空格分割成数值
    $vals = expl($val);
    $a = evev($vals[0]);
    $b = evev($vals[1]);
    $ao = evev($vals[2]);
    $bo = evev($vals[3]);

    //已知条件
    $know = array();
    array_push($know, $val);
    $v0 = eveq($vals[0]);
    array_push($know, "a边长:$v0");
    $v1 = eveq($vals[1]);
    array_push($know, "b边长:$v1");
    $v2 = eveq($vals[2]);
    array_push($know, "{$math['ang']}a:$v2{$math['o']}");
    $v2 = eveq($vals[3]);
    array_push($know, "{$math['ang']}b:$v2{$math['o']}");

    //计算步骤
    $step = array();
    array_push($step, "求三角形的c边:");
    array_push($step, "{$math['eq']} $a{$math['mul']}cos{$bo}{$math['o']}{$math['add']}$b{$math['mul']}cos{$ao}{$math['o']}");
    $cosbo = cos(deg2rad($bo));
    $cosao = cos(deg2rad($ao));
    array_push($step, "{$math['eq']} $a{$math['mul']}$cosbo{$math['add']}$b{$math['mul']}$cosao");
    $cosboa = bcmul($a, $cosbo);
    $cosaob = bcmul($b, $cosao);
    array_push($step, "{$math['eq']} $cosboa{$math['add']}$cosaob");
    $c = bcadd($cosboa, $cosaob);
    array_push($step, "{$math['eq']} $c");

    array_push($step, "求三角形的{$math['ang']}C:");
    array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$a{$math['sup2']}{$math['add']}$b{$math['sup2']}{$math['sub']}$c{$math['sup2']}{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$a{$math['mul']}$b{$math['rpar']}{$math['rsqb']}");
    $aa = bcmul($a, $a);
    $bb = bcmul($b, $b);
    $cc = bcmul($c, $c);
    array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$aa{$math['add']}$bb{$math['sub']}$cc{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$a{$math['mul']}$b{$math['rpar']}{$math['rsqb']}");
    $aabb = bcadd($aa, $bb);
    $ab = bcmul($a, $b);
    array_push($step, "{$math['eq']} arccos{$math['lsqb']}{$math['lpar']}$aabb{$math['sub']}$cc{$math['rpar']}{$math['div']}{$math['lpar']}2{$math['mul']}$ab{$math['rpar']}{$math['rsqb']}");
    $aabbcc = bcsub($aabb, $cc);
    $ab2 = bcmul($ab ,2);
    array_push($step, "{$math['eq']} arccos{$math['lpar']}$aabbcc{$math['div']}$ab2{$math['rpar']}");
    $aabbccab2 = $aabbcc/$ab2;
    array_push($step, "{$math['eq']} arccos{$math['lpar']}$aabbccab2{$math['rpar']}");
    $acosaabbccab2= acos($aabbccab2);
    array_push($step, "{$math['eq']} 弧度值:{$acosaabbccab2}");
    $co = rad2deg($acosaabbccab2);
    array_push($step, "{$math['eq']} 角度值:{$co}{$math['o']}");

    //算出结果
    $ends = array();
    array_push($ends, "三角形的a边:{$a}");
    array_push($ends, "三角形的b边:{$b}");
    array_push($ends, "三角形的c边:{$c}");
    $aoo = round($ao,1);
    $boo = round($bo,1);
    $coo = round($co,1);
    array_push($ends, "三角形的{$math['ang']}A:{$aoo}{$math['o']}");
    array_push($ends, "三角形的{$math['ang']}B:{$boo}{$math['o']}");
    array_push($ends, "三角形的{$math['ang']}C:{$coo}{$math['o']}");

    //公式表示
    $home = array();
    array_push($home, "三角形的c边:c{$math['eq']}a{$math['mul']}cos{$math['ang']}B{$math['add']}b{$math['mul']}cos{$math['ang']}A");
    array_push($home, "三角形的{$math['ang']}C:{$math['ang']}C{$math['eq']}arccos{$math['lsqb']}{$math['lpar']}a{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}c{$math['sup2']}{$math['rpar']}{$math['sol']}2ab{$math['rsqb']}");

    //解释说明
    $info = array();
    array_push($info, "三角函数的余弦定理,可求任意三角形");
    array_push($info, "已知任意三角形的两边及两边所夹的角,求第三边:");
    array_push($info, "a{$math['sup2']}{$math['eq']}b{$math['sup2']}{$math['add']}c{$math['sup2']}{$math['sub']}2bc{$math['mul']}cos{$math['ang']}A");
    array_push($info, "b{$math['sup2']}{$math['eq']}a{$math['sup2']}{$math['add']}c{$math['sup2']}{$math['sub']}2ac{$math['mul']}cos{$math['ang']}B");
    array_push($info, "c{$math['sup2']}{$math['eq']}a{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}2ab{$math['mul']}cos{$math['ang']}C");

    array_push($info, "已知三角形的两边及与第三边所夹的两角,求第三边:");
    array_push($info, "a{$math['eq']}b{$math['mul']}cos{$math['ang']}C{$math['add']}c{$math['mul']}cos{$math['ang']}B");
    array_push($info, "b{$math['eq']}c{$math['mul']}cos{$math['ang']}A{$math['add']}a{$math['mul']}cos{$math['ang']}C");
    array_push($info, "c{$math['eq']}a{$math['mul']}cos{$math['ang']}B{$math['add']}b{$math['mul']}cos{$math['ang']}A");

    array_push($info, "已知三角形的三边,求角度数:");
    array_push($info, "cos{$math['ang']}A{$math['eq']}{$math['lpar']}c{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}a{$math['sup2']}{$math['rpar']}{$math['sol']}2bc");
    array_push($info, "cos{$math['ang']}B{$math['eq']}{$math['lpar']}a{$math['sup2']}{$math['add']}c{$math['sup2']}{$math['sub']}b{$math['sup2']}{$math['rpar']}{$math['sol']}2ac");
    array_push($info, "cos{$math['ang']}C{$math['eq']}{$math['lpar']}a{$math['sup2']}{$math['add']}b{$math['sup2']}{$math['sub']}c{$math['sup2']}{$math['rpar']}{$math['sol']}2ab");

    know($know);
    ends($ends);
    home($home);
    step($step);
    info($info);
}
?>
<?php include ROOT.'/assets/php/foot.php'; ?>

结果:

☁参考上图[2022-12-04]  

☀求三角形各边各角度已知两边和第三边所夹两角  
30 40 25 35  
a边长:30  
b边长:40  
∠a:25°  
∠b:35°  

♠算出结果  
三角形的a边:30  
三角形的b边:40  
三角形的c边:60.82  
三角形的∠A:25°  
三角形的∠B:35°  
三角形的∠C:120°  

♦公式表示  
三角形的c边:c=a×cos∠B+b×cos∠A  
三角形的∠C:∠C=arccos[(a²+b²−c²)/2ab]  

♣计算步骤  
求三角形的c边:  
= 30×cos35°+40×cos25°  
= 30×0.81915204428899+40×0.90630778703665  
= 24.57+36.25  
= 60.82  
求三角形的∠C:  
= arccos[(30²+40²−60.82²)÷(2×30×40)]  
= arccos[(900.00+1600.00−3699.07)÷(2×30×40)]  
= arccos[(2500.00−3699.07)÷(2×1200.00)]  
= arccos(-1199.07÷2400.00)  
= arccos(-0.4996125)  
= 弧度值:2.0939477136999  
= 角度值:119.97436651607°  

♥解释说明  
三角函数的余弦定理,可求任意三角形  
已知任意三角形的两边及两边所夹的角,求第三边:  
a²=b²+c²−2bc×cos∠A  
b²=a²+c²−2ac×cos∠B  
c²=a²+b²−2ab×cos∠C  
已知三角形的两边及与第三边所夹的两角,求第三边:  
a=b×cos∠C+c×cos∠B  
b=c×cos∠A+a×cos∠C  
c=a×cos∠B+b×cos∠A  
已知三角形的三边,求角度数:  
cos∠A=(c²+b²−a²)/2bc  
cos∠B=(a²+c²−b²)/2ac  
cos∠C=(a²+b²−c²)/2ab  

标签:cos,30,40,arccos,已知,各边,100027,三角形,两角
From: https://www.cnblogs.com/onestopweb/p/16950302.html

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